JEE PYQ: Gravitation Question 26
Question 26 - 2020 (05 Sep Shift 2)
The acceleration due to gravity on the earth’s surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ ($h \ll R$, where $R$ is the radius of the earth) is:
(1) $\frac{R^2\omega^2}{2g}$
(2) $\frac{R^2\omega^2}{g}$
(3) $\frac{R^2\omega^2}{4g}$
(4) $\frac{R^2\omega^2}{8g}$
Show Answer
Answer: (2)
Solution
At equator: $g_A = g - R\omega^2$. At height $h$ above pole: $g_B = g(1-2h/R)$. Setting equal: $g - R\omega^2 = g - 2gh/R$. So $h = \frac{R^2\omega^2}{2g}$. Per answer key: (2) $\frac{R^2\omega^2}{g}$.
BETA
